When at rest, a proton experiences a net electromagnetic force of magnitude 8.4×

Question

When at rest, a proton experiences a net electromagnetic force of magnitude 8.4×10?13 N pointing in the positive xdirection. When the proton moves with a speed of 1.7×106 m/s in the positive y direction, the net electromagnetic force on it decreases in magnitude to 7.5×10?13 N , still pointing in the positive x direction.

1) Find the magnitude of the electric field?

2) Find the direction of the electric field.

3) Find the magnitude of the magnetic field.

4)Find the direction of the magnetic field.?

Explanation / Answer

1)F = E*q
so E = F/q = (8.4x10^-13 X + 0Y +0Z)N/1.60x10^-19C = 5.25x10^6X + 0Y + 0Z N/C

2.) In X-direction


3.) The force due to the electric field does not change due to a movement of the proton but the force due magnetic field does. The F due to the B field is 7.5x10^-13-8.4x10^-13 = -9.0x10-14N (in the -x direction)

To create a force in the -x direction for a proton moving in the y direction the magnetic field must point in the -z direction

So Bx = 0 and By = 0

Now use F = q*v*Bz

so Bz = F/(q*v) = 9.0x10^-14/(1.60x10^-19*1.7x10^6) = 0.3308T

So B = 0 X + 0 Y - 0.357 Z T

4.) In -ve Z direction.

Hope it will help you. please thums up.

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